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rjkz [21]
3 years ago
14

PLS HELP. ITS FOR AN ASSIGNMENT DUE SOON!!! IF YOU DONT KNOW THE ANSWER PLEASE DONT SAY ANYTHING.

Mathematics
1 answer:
AfilCa [17]3 years ago
4 0
Answer:
The numbers are 9, 7, 4, 11
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1. Geoff rode his bike along an 8-mile path and lost his cell phone at some random location
Elina [12.6K]

Answer:

0.3125

Step-by-step explanation:

Use definition of geometric probability:

P=\dfrac{\text{Desired Length}}{\text{Total Length}}

In your case,

Total Length = 8 miles

Desired Length = 7 - 4.5 = 2.5 miles,

so the probability is

P=\dfrac{2.5}{8}=\dfrac{25}{80}=\dfrac{5}{16}=0.3125

5 0
4 years ago
3(x-2) <br> I don’t know how to do this
grandymaker [24]

Answer:

3x - 6 should be it

6 0
2 years ago
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Verify whether the indicated numbers are zeroes of their corresponding polynomials.
natita [175]

Answer:

  • No, incorrect

Step-by-step explanation:

<u>Given</u>

  • Q(s) = -4s^3 + 7s^2 - 24;
  • s = -4 and 1

<u>Verifying zeroes</u>

Q(-4) =

  • -4(-4)^3 + 7(-4)^2 - 24 =
  • 256 + 112 - 24 =
  • 344
  • Incorrect as 344 ≠ 0

Q(1) =

  • -4(1)^3 + 7(1)^2 - 24 =
  • -12 + 7 - 24 =
  • -29
  • Incorrect as -29 ≠ 0

8 0
3 years ago
Find the missing factor.<br> 200 x ____ = 600<br> A) 2<br> B) 3<br> C) 4<br> D) 5
kykrilka [37]

Answer:

B. 3

Step-by-step explanation:

we know 2x3 = 6 soo just add 2 zeros = 600

4 0
2 years ago
Read 2 more answers
Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
NeX [460]

Answer:

(a) 0.343

(b) 0.657

(c) 0.189

(d) 0.216

(e) 0.353

Step-by-step explanation:

Let P(a vehicle passing the test) = p

                        p = \frac{70}{100} = 0.7  

Let P(a vehicle not passing the test) = q

                         q = 1 - p

                         q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

                           = 0.7 × 0.7 × 0.7

                           = 0.343

(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

      = (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)

      = 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

      = 0.657

(c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

                 = 0.063 + 0.063 + 0.063

                 = 0.189

(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

                 = 0.063 + 0.063 + 0.063 + 0.027

                 = 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

=  (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

                  = 0.973  

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

                         = \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}

                         = \frac{0.343}{0.973}

                         = 0.353

3 0
4 years ago
Read 2 more answers
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